1

Additional Mathematics Marking Scheme - Paper 2

No Solution and Mark Scheme Sub

Marks Total

Marks

1

2 (a)

(b)

yx 22 or 2

2 xy

7)22()22( 22 yyyy or

72

2

2

22

2

xxxx

0363 2 yy

0122 yy or

0243 2 x

082 x

Solve the quadratic equation

)1(2

)1)(1(4)2()2( 2 y or

)1(2

)8)(1(4)0(0 2 x

𝑥 = −2.82, 𝑥 = 2.82

𝑦 = 2.41, 𝑦 = −0.41

a = 42 or d = 4 OR by listing 42,46,50,…

4)16(426 T OR by listing 42, 46, 50, 54, 58, 62

= 62

10000 /400 = 25 pusingan

4)125()42(22

2525 S or by listing.

= 2250

P1

K1

K1

N1

N1

P1

K1

N1

P1

K1

N1

5

3

3

6

2

No Solution and Mark Scheme Sub

Marks Total

Marks

3 (a)

( b)

(c)

𝑘𝑥 − 3 = 0

𝑘 = 2

𝑓−1(𝑥) = 3𝑥

∗2𝑥−5 or

−3𝑥

5− ∗2𝑥

OR 5𝑥

∗2𝑥−3= −2

3 − 4 (3𝑥

2𝑥−5 )

2

3

𝑔𝑓−1(−2) = 1

3

𝑥 =3 − 𝑦

4

ℎ(𝑦) = 12 (3 − 𝑦

4) + 23

ℎ(𝑥) = 32 − 3𝑥

K1

N1

K1

K1

N1

K1

K1

N1

2

3

3

8

4 (a)

(b)

Seen 40.5 or 14 or 6 P1

𝑚 = 40.5 + [1

2(32 )−14

6] 10 K1

= 43.83 N1

seen 68 818 (tunjuk cara)

min = 1426

32

σ = √68 818

32− (

1426

32) ²

= 12.84

P1

K1

N1

K1

K1

K1

N1

3

4

7

3

No. Solution and Mark

Scheme

Sub

Marks

Total

Marks

5 (a)

(b)

𝑑𝑦

𝑑𝑥= 1 − 𝑥−3

1 − 𝑥−3 = 0

𝑥 = 1

𝑦 = 1 + 1

2(1)2

(1,3

2)

𝑉 = 64ℎ

𝑑𝑉

𝑑ℎ= 64, or

𝑑ℎ

𝑑𝑡= 0.6

𝑑𝑉

𝑑𝑡=

𝑑𝑉

𝑑ℎ𝘹

𝑑ℎ

𝑑𝑡

= 64 𝘹 0.6

= 38.4

K1

K1

N1

K1

K1

K1

N1

3

4

7

4

No. Solution and Mark Scheme Sub

Marks

Total

Marks

6. (a)

(b)

(c)

When y = 0; M(3, 0)

When x = 0; N(0, -6)

2,2

21

)6(1)0(2

21

)0(1)3(2

P

or

𝐿𝑢𝑎𝑠 ∆𝑁𝑂𝑃 = 1

2|(12) ∗ −(0)|

= 6

2 × 𝑚2= − 1

𝑚2 = − 1

2

𝑦 − (−2) = −1

2 (𝑥 − 2)

𝑦 = −1

2𝑥 − 1

K1

N1

K1

N1

K1

K1

N1

2

2

3

7

5

No.

Solution and Mark Scheme

Sub

Marks

Total

Marks

7 (a)(i)

(ii)

(b) (i)

(ii)

𝑃(𝑋 = 7) = 9C7 (0.7)7 (0.3)2

= 0.2668

P(X = 7) + P(X = 8) + P(X= 9)

= 0.4628

K1

N1

K1

N1

K1

N1

K1

N1

K1

N1

2

2

2

4

10

9938.0

]5.2[

]12

160130[

]130[

ZP

P

XP

9.149

842.012

160

842.0

2000.0)(

k

k

Z

kXP

6

No. Solution and Mark Scheme Sub

Marks

Total

Marks

8(a)

(b)

(c)

65

16)cos(

5

3

13

12

5

4

13

5)cos(

5

3sin,

13

5cos

sinsincoscos)cos(

QP

QP

QandP

QPQPQP

xxy 0;4sin4

x

-2 y= - 2

-4

Shape of sine graph

Shape of negative sine curve

2 cycles

Amplitude = 4

2y

Sketch the straight line y = - 2 correctly

Numbers of solutions = 4

P1

K1

N1

P1

P1

P1

P1

N1

K1

N1

3

4

3

10

π

4 •

•

•

•

•

•

•

•

•

7

No Solution and Mark Scheme

Sub

Marks

Total

Marks

10 (a)

(b)

(c)

i) yxPM 63

ii) )(3

2NMONOS

yx 410

OS )2(4)3(10 jii

34.93

)63( yxhPT

)45( yxkTS

ii) PS 5

4yxx 410)15(

)4)5()63( ykxkyhxh

7

2h

7

4k

N1

K1

N1

K1

N1

N1

N1

K1

N1

N1

3

2

5

10

8

No Solution and Mark Scheme Sub

Marks

Total

Marks

11 (a)

(b)

(c)

xdx

dy6 ,

y = ∫ 6𝑥 𝑑𝑥

y = 3x² + c

6 = 3(1)2 + 𝑐

𝑦 = 3𝑥² + 3

dxxArea 1

0

2 33)61(

= 103 36 xx

= 6 – 4

= 2

dyy

Volume

6

3 3

3

=

6

3

2

6

y

y

=

6

90

= 5.12

3or

K1

K1

N1

K1K1

K1

N1

K1

K1

N1

3

4

3

10

9

No. Solution and Mark Scheme

Sub

Marks

Total

Marks

12(a)

(b)

(c)

(d)

𝑣 = 8𝑡 − 2𝑡2 − 6 8 − 4𝑡 = 0

𝑡 = 2

𝑣 = 8(2) − 2(2)² − 6 = 2

𝑣 = 8𝑡 − 2𝑡2 − 6 = 0 −𝑡2 + 4𝑡 − 3 = 0 (𝑡 − 3)(𝑡 − 1) = 0

𝑡 = 3, 𝑡 = 1

𝑠 = −2

3𝑡3 + 4𝑡2 − 6𝑡

|−2

3𝑡3 + 4𝑡2 − 6𝑡|

0

1 + |−

2

3𝑡3 + 4𝑡2 − 6𝑡|

1

3

= 16

3

1

1

1

1

1

1

1

1

1

1

4

2

2

2

10

2

1 3

-6

10

No Solution and Mark Scheme Sub

Marks Total

Marks

13 (a)

(b)

(c)

(d)

8² = 9² + 5.6² − 2(9)(5.6) cos BCD BCD = 61.33º or 61º 20’

oCAE 33.61sin

2.11

sin

12

CAE = 70.06o

AEC = 48.61o

o

AC

33.61sin

2.11

61.48sin

AC = 9.58 cm

Area BDC = 33.61sin96.52

1 and

Area EBC = 33.61sin126.52

1 OR

29.48 – 22.11

Area EBC - Area EBC OR

33.61sin96.52

1

3

1

Area BDE = 7.37 cm2

K1

N1

K1

N1

N1

K1

N1

K1

K1

N1

2

3

2

3

10

11

No

Solution and Mark Scheme Sub

Marks Total

Marks

14 (a)

(b)

(c) (i)

(ii)

𝑥 + 𝑦 ≤ 400

𝑥 ≤ 3𝑦 or 𝑦 ≥1

3

80𝑥 + 60𝑦 ≥ 7200

Graph 1 correct line

All correct line

Correct shaded region

300,100) 80(300) +60(100)

= 30 000

50 orang

N1

N1

N1

K1

N1

N1

N1

K1

N1

N1

3

3

4

10

12

No Solution and Mark Scheme Sub

Marks Total

Marks

15 (a)

(b)

(c)

(d)

x 6.00

𝑦 = 110

𝐼 =120(7) + 115(3) + 110(6) + 125(8)

24

118.54

115

𝑃2012 × 100 = 118.54

𝑃2012 = 97.01

108(7) + 120.75(3) + 115.5(6) + 125(8)

24

= 117.14

N1

N1

K2

N1

K1

N1

K2

N1

2

3

2

3

10